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Show that the Kelvin-Planck and the Clausius expressions of the second law are equivalent.

Short Answer

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Answer: Yes, the Kelvin-Planck and Clausius statements of the second law of thermodynamics are equivalent. These statements describe the limitations and restrictions in heat transfer processes, with the Kelvin-Planck statement relating to heat engine efficiency, and the Clausius statement relating to the spontaneous flow of heat between two bodies. Violating one statement results in the violation of the other, proving their equivalence.

Step by step solution

01

Kelvin-Planck Statement of Second Law

The Kelvin-Planck statement of the second law of thermodynamics is stated as follows: No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of this heat into work. In simpler terms, it states that a heat engine cannot have 100% efficiency.
02

Clausius Statement of Second Law

The Clausius statement of the second law of thermodynamics is stated as follows: No process is possible whose sole result is the transfer of heat from a cooler body to a hotter body. In simpler terms, it states that heat cannot spontaneously flow from a colder body to a hotter body.
03

Establish the Connection between Both Statements

First, we have to identify the connection between heat engines (Kelvin-Planck) and spontaneous heat transfer (Clausius). Both statements relate to the impossibility of certain processes involving heat. The Kelvin-Planck statement deals with heat engines and their inability to be completely efficient, while the Clausius statement deals with the spontaneous flow of heat between two bodies in thermal contact. The connection here is that both processes involve the transfer of heat and energy, with certain limitations.
04

Demonstrate the Equivalence

To show that the Kelvin-Planck and Clausius statements are equivalent, we have to demonstrate that the violation of one statement would also violate the other statement. To do this, we assume a hypothetical process that violates one of the statements and then show that this process would also violate the other statement. 1. Assume the Kelvin-Planck statement is violated: Let's assume that there exists a heat engine that can completely convert absorbed heat into work, thus violating the Kelvin-Planck statement. With this hypothetical engine, we can transfer heat from a cooler body to a hotter body, using the work output obtained from the complete conversion of heat. This process will violate the Clausius statement, as heat is being transferred spontaneously from a cooler body to a hotter body. 2. Assume the Clausius statement is violated: Let's assume that there exists a spontaneous process that can transfer heat from a cooler body to a hotter body, thus violating the Clausius statement. With this hypothetical process, we can generate work by transferring heat from a cooler body to a hotter body and use this heat difference as a source for a heat engine. Since we have assumed the heat transfer as spontaneous, this would imply that the heat engine is absorbing heat and converting all of it into work, thereby violating the Kelvin-Planck statement.
05

Conclusion

Since we have shown that the violation of one statement would also violate the other statement, we can conclude that the Kelvin-Planck and Clausius statements of the second law of thermodynamics are equivalent. Both statements provide a clear description of the limitations and restrictions associated with heat transfer processes and together describe the fundamental principles of the second law of thermodynamics.

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Most popular questions from this chapter

It is well known that the thermal efficiency of heat engines increases as the temperature of the energy source increases. In an attempt to improve the efficiency of a power plant, somebody suggests transferring heat from the available energy source to a higher-temperature medium by a heat pump before energy is supplied to the power plant. What do you think of this suggestion? Explain.

A Carnot heat engine receives heat from a reservoir at \(1700^{\circ} \mathrm{F}\) at a rate of \(700 \mathrm{Btu} / \mathrm{min}\) and rejects the waste heat to the ambient air at \(80^{\circ} \mathrm{F}\). The entire work output of the heat engine is used to drive a refrigerator that removes heat from the refrigerated space at \(20^{\circ} \mathrm{F}\) and transfers it to the same ambicnt air at \(80^{\circ} \mathrm{F}\). Determine \((a)\) the maximum rate of heat removal from the refrigerated space and ( \(b\) ) the total rate of heat rejection to the ambient air.

A typical electric water heater has an efficiency of 95 percent and costs \(\$ 350\) a year to operate at a unit cost of electricity of \(\$ 0.11 / \mathrm{kWh}\). A typical heat pump-powered water heater has a COP of 3.3 but costs about \(\$ 800\) more to install. Determine how many years it will take for the heat pump water heater to pay for its cost differential from the energy it saves.

Why is it important to clean the condenser coils of a household refrigerator a few times a year? Also, why is it important not to block airflow through the condenser coils?

A refrigeration system uses water-cooled condenser for rejecting the waste heat. The system absorbs heat from a space at \(25^{\circ} \mathrm{F}\) at a rate of \(24,000 \mathrm{Btu} / \mathrm{h}\). Water enters the condenser at \(65^{\circ} \mathrm{F}\) at a rate of \(1.45 \mathrm{lbm} / \mathrm{s}\). The COP of the system is estimated to be \(1.9 .\) Determine \((a)\) the power input to the system, in \(\mathrm{kW},(b)\) the temperature of the water at the exit of the condenser, in \(^{\circ} \mathrm{F}\) and \((c)\) the maximum possible COP of the system. The specific heat of water is \(1.0 \mathrm{Btu} / \mathrm{bm} \cdot^{\circ} \mathrm{F}\)

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